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I'm trying to understand the following theorem of the book Banach space theory of Fabian et. al.:

enter image description here

I'm having troubles to see why is it true what I've marked in yellow.

As I see it, we have that $U$ is $\mathcal{T}$-closed, convex and balanced. So, the bipolar theorem applied with respect to the dual pair $\langle E, F \rangle$ tells us that

$$U^{\circ\circ} = \overline{\text{conv}\left( U \cup \{0_E\} \right)}^{w(E,F)}$$

As $U$ is convex and balanced we can simplify it to

$$U^{\circ\circ} = \overline{U}^{w(E,F)}$$

Now, I don't see how to get $U$ from here, because we know that $U$ is $\mathcal{T}$-closed, but that doesn't implie that it is $w(E,F)$-closed.

What am I missing?

Thanks in advance for the replies.

Eparoh
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1 Answers1

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One way is to use the following fact.

Proposition. Let $(E, \mathcal T)$ be a locally convex space and let $E'$ denote its continuous dual. Then a convex set $A \subset E$ is $w(E, E')$-closed if and only if it is $\mathcal T$-closed.

Proof. Since $\mathcal T$ is stronger than $w(E,E')$, $w(E,E')$-closedness implies $\mathcal T$-closedness (without mentioning convexity of $A$). Now assume that $A$ is $\mathcal T$-closed and convex. Then for all $x \notin A$ there exists $f \in E'$ such that $f(x) \notin \overline{f(A)}$ by the separation theorem (a simple corollary of Hahn-Banach theorem). That is, there exists $\varepsilon > 0$ such that $V = \{y \in E: |f(y) - f(x)| < \varepsilon\}$ does not intersect $A$. By definition $V$ is a $w(E,E')$-neighborhood of $x$ that does not intersect $A$. Thus, $E \setminus A$ is $w(E,E')$-open. $\blacksquare$

Matsmir
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